a boy P stands on top of tower.another boy Q throws a ball vertically up from the ground so that it just reaches P & time taken is t₀..the speed with which the ball must be thown so that it reaches ground in time t₀/2 is

To solve the problem of how fast boy Q needs to throw the ball so that it reaches the ground in half the time it takes to reach boy P at the top of the tower, we can break it down into a few logical steps. Let's analyze the motion of the ball in two parts: the upward motion to reach P and the downward motion to return to the ground.

Understanding the Motion of the Ball

When boy Q throws the ball vertically upward, it travels against the force of gravity until it reaches the height of boy P. The time taken for this upward journey is given as \( t_0 \). The ball will then take the same amount of time, \( t_0 \), to fall back down to the ground from the height of P, making the total time for the round trip \( 2t_0 \).

Calculating the Initial Velocity

We can use the equations of motion to find the initial velocity required for the ball to reach boy P. The formula for the maximum height \( h \) reached by an object thrown upwards is given by:

• \( h = v_i t - \frac{1}{2} g t^2 \)

Where:

• \( v_i \) is the initial velocity of the ball.
• \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
• \( t \) is the time taken to reach the maximum height (which is \( t_0 \) in this case).

At the peak height, the final velocity is zero. Thus, we can rearrange the equation to find the initial velocity:

• \( 0 = v_i - g t_0 \)
• Therefore, \( v_i = g t_0 \)

Finding the Speed for Half the Time

Now, we need to determine the speed with which the ball must be thrown so that it reaches the ground in half the time, \( \frac{t_0}{2} \). For this downward journey, we can use the same equation of motion, but we need to consider the time taken to fall from the height of P back to the ground:

• \( h = \frac{1}{2} g t^2 \)

Substituting \( t = \frac{t_0}{2} \) into the equation gives us:

• \( h = \frac{1}{2} g \left(\frac{t_0}{2}\right)^2 = \frac{1}{2} g \frac{t_0^2}{4} = \frac{g t_0^2}{8} \)

Since we already established that the height \( h \) is also equal to \( g t_0^2 \) (from the upward motion), we can set these two expressions for height equal to each other:

• \( g t_0^2 = \frac{g t_0^2}{8} \)

This means that the ball must be thrown with an initial velocity that allows it to reach this height in half the time. To find the required initial velocity for this downward motion, we can rearrange the equation:

• \( v_f = v_i - g t \)
• Where \( v_f \) is the final velocity just before hitting the ground, which is equal to \( g t_0/2 \) when it falls from height \( h \).

Thus, the initial velocity \( v_i \) needed for the ball to reach the ground in \( \frac{t_0}{2} \) can be calculated as:

• \( v_i = g \left(\frac{t_0}{2}\right) \)

Final Thoughts

In summary, boy Q must throw the ball with an initial speed of \( g \frac{t_0}{2} \) to ensure it reaches the ground in half the time it takes to reach boy P. This approach illustrates the relationship between time, velocity, and the effects of gravity on an object in motion. Understanding these principles can help you tackle similar problems in physics effectively.